Question

In a survey of $500$ TV viewers: $285$ watch football (F), $195$ hockey (H), $115$ basketball (B); $45$ watch F&B, $70$ watch F&H, $50$ watch H&B, and $50$ watch none. How many watch exactly one of the three games? 

• A. 325
• B. 405
• C. 310
• D. 372

Hint:

For "exactly one", compute each group as:  
\[\text{Only }F = F - (F \cap H + F \cap B) + t\]  and sum; find \(t\) via inclusion–exclusion.
 

Answer 1

The Correct Option is A

Watching at least one: \(|F\cup H\cup B|=500-50=450.\) Using inclusion-exclusion with $t=|F\cap H\cap B|$: \[ 450=285+195+115-(45+70+50)+t=430+t \Rightarrow t=20. \] Only-F $=285-(70+45)+20=190$, 
Only-H $=195-(70+50)+20=95$, 
Only-B $=115-(45+50)+20=40$. 
Exactly one $=190+95+40=\boxed{325}$. 

Answer 2

Instead of computing each of only-football, only-hockey, and only-basketball separately, work out the counts of exactly-two and exactly-three viewers first, then subtract these from the total who watch at least one game.

At least one game: \( 500-50=450 \) (since \( 50 \) watch none). By inclusion-exclusion, with \( t=|F\cap H\cap B| \): \[ 450=285+195+115-(45+70+50)+t=430+t\ \Rightarrow\ t=20. \]

The sum of the three pairwise-intersection counts, \( 45+70+50=165 \), counts each exactly-two viewer once and each exactly-three viewer three times: \[ (\text{exactly two})+3t=165\ \Rightarrow\ (\text{exactly two})=165-3(20)=105. \]

1. Option A (\( 325 \)): Exactly one \( = \) at least one \( - \) exactly two \( - \) exactly three \( =450-105-20=325 \). This matches.
2. Option B (\( 405 \)): This does not equal \( 450-105-20 \); rejected.
3. Option C (\( 310 \)): This also does not match the computed value; rejected.
4. Option D (\( 372 \)): This is likewise inconsistent with the computed breakdown; rejected.

Breaking the total into exactly-two and exactly-three viewers before subtracting confirms that exactly \( 325 \) people watch only one of the three games.

Hence, the correct answer is 325.